How Do You Calculate Flux Through a Box with Varying Electric Fields?

[Jim Newt]

Hi all,

I hope you can see the attachment. Could someone give me a more detailed explanantion of solving this problem? Any help would be greatly appreciated. Thanks,

Jim

 

[AstroRoyale]

The answer looks right to me, from a quick glance. What are you having problems with?

 

[Jim Newt]

I'm having problems with the whole concept. For instance, why does the "right side" have a value of 18 dot 9, and the bottom have a value of -4 dot -9? Thanks,

Jim

 

[AstroRoyale]

Well, we have some field that is pointing through the surfaces of the box. So you are asked to find the flux through the entire box. The box obviously has six faces, I'll go through a few of them. . .

It's easier to write $$dA=\hat{n}\cdot dA$$, more intuitive to me at least . . .

top: the normal vector $$\hat{n}$$ points in the positive direction, so the flux
is $$\int \vec{E} \cdot \hat{n}dA= \int +E_{z} dA = -4\int dA = -36$$

Bottom: the normal vector $$\hat{n}$$ to the bottom points in the -z direction, so the flux is is
$$\int \vec{E} \cdot \hat{n}dA= \int -E_{z} dA = 4\int dA = +36$$

Right facing side: the normal vector $$\hat{n}$$ to the right points in the +j direction, so the flux is
$$\int \vec{E} \cdot \hat{n}dA= \int E_{y} dA = \int 2y^2 dA = \int 2y^2 dx dz =2y^2_{y=3} \int dxdz = 18 * 9 = 162$$

left facing side: Same stuff here, but y=0 on the left side so the flux=0

I'll let you do the front and back sides. They are similar to the right/left faces. Hope that helps. . .